What do you do in a quadratic equation if there is a number before the x^2?

Question

One example I can give is:
2x^2 * x * 6

let me just state that * is a multiplycation symbol

Answer 1

hi,

it does not matter if there is a number before the x^2 term or not. it works the same either way.

if the poly can be factored, you do that; you just need to use the factors of the number in front of x^2 to do that. and if it can not be factored, you would use the quadratic formula where the number in front of x^2 becomes the variable a in the quadratic formula.

so, for example,
2x^2 -3x - 2 can be factored as
(2x+1) ( x-2)

and that's all there is to it.

Answer 2

It is difficult to make out what your question is.
You mention an "equation" but an equation requires an = sign and there is no = sign in the question.
Is it (2x²) X (x) X (6) where X is multiplication symbol?
If so answer is 2 X x X x X x X 6
= 6 X 2 X x X x X x
= 12 x ³
Looking at all the different answers, it is clear that the wording of the question is giving us all cause for concern!

Answer 3

You remedy it via splitting the midsection term in accordance to the consistent. the two numbers could have product 10 and the sum could be 7 ( coeff of x). So here you have x^2 +5x+2x+10=0 (the midsection term 7x has been chop up into 5x +2x) now you're taking the easy factors interior the words x(x+5) +2(x+5)=0 (x+2)(x+5)=0 now equate the two factors to 0 and get the values of x x+2=0 and x+5=0 so x= -2 and x= -5

Answer 4

I'm assuming what you are doing is factoring.
I'll do 2x^2 + 7x + 6 as an example.

There are a couple different methods you can use. The one I suggest is this:

1.) Multiply the coefficient of the x^2 term and the constant...in this case, 2*6=12.

2.) Find factors of this product that add to give you the coefficient of the middle term. 12 = 3*4, 3+4=7, so those are the two numbers we want.

3.) Split the middle term into two pieces, with those two numbers being the coefficients. In this example, 7x = 3x + 4x. So the problem is now 2x^2 + 3x + 4x + 6.

4.) Solve by factoring:

2x^2 + 3x + 4x + 6
x(2x+3) +2 (2x+3)
(x+2)(2x+3)

The example you gave wouldn't factor, but a similiar problem 2x^2 - x + 6 would.

Answer 5

First of all, are you sure it's 2x^2 * x * 6? Not 2x^2 + x + 6? Because it wouldn't be a quadratic if they multiply.

To solve a quadratic equation, the basic formula is:
x = [-b +or- sq. root(b^2 - 4ac)]/2a
Where a = co-efficient of x^2
b = co-efficient of x
c = the last single number

Just sub the numbers in, and you should be fine. One case though, if the number inside the square root is less than 0, you will need to use imaginary number "i" (which is in the case of "2x^2 + x + 6"), but if you havn't learnt that, it will be left as x = undefined.

Answer 6

you have to multiply that number with the last number
for eg:
2x^2*x*6=0
2*6=12(product)
1 is the sum
you should find two numbers that when added or subtracted should give you 2 and ehen multiplied gives you 12
the no. for this eq is 4and -3 and then solve

Answer 7

I think you mean 2x^2+x+6--right. Well, you just let a=2 and do everything the same: [-1+-sqrt(1-48)]/4.

Answer 8

it is not a quadratic equation form.

it becomes 12 x^3 now. Quadratics have x^2 .

Answer 9

im thinking you * should be + because the quadradic formula is used when you have and equation in the form

ax^2+bx+c=0

b+- [{sqrt}(b^2-4ac)] / 2a

so if there is something before the x^2 you plug it in as the a in the quadradic formula